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Assume I can express a rank-deficient, $N\times N$ symmetric covariance matrix $\Sigma$ as

\begin{equation} \Sigma=\mathbf{USU}^\top \end{equation}

where $\mathbf{U}$ is an $L\times N$ orthonormal matrix, i.e. $\mathbf{U^\top U}=\mathbf{I}_L$.

I am wondering whether the Frobenius norm $|\Sigma|$ is equal to $|\mathbf{S}|$. I can see how this is true for the special case where $\mathbf{U}$ is an orthonormalizing (eigenvector) basis (so that $\mathbf{S}$ is diagonal), and haven't been able to show otherwise in numerical experiments, but I'm not sure how to evaluate more generally.

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The answer is yes. Note that $$ \begin{align} |\Sigma|^2 &= |USU^T|^2 = \operatorname{tr}[(USU^T)(USU^T)^T] \\ & = \operatorname{tr}[US(U^TU)S^TU^T] \\ & = \operatorname{tr}[U(SS^TU^T)] = \operatorname{tr}[(SS^TU^T)U] \\ & = \operatorname{tr}[SS^T(U^TU)] = \operatorname{tr}[SS^T] = |S|^2. \end{align} $$

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Alternatively, complete $U$ to a square orthogonal matrix $V=\pmatrix{U&\ast}$. Then $$ \Sigma=USU^T=V\pmatrix{S\\ &0}V^T. $$ Therefore $\|\Sigma\|_F =\left\|V\pmatrix{S\\ &0}V^T\right\|_F =\left\|\pmatrix{S\\ &0}\right\|_F=\|S\|_F$.

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